how many ml of 0.5 M BaCl2 are needed to have 75 g of BaCl2?
mols = M x L = ?
idl
720
To determine the volume of 0.5 M BaCl2 needed to have 75 g of BaCl2, we need to first calculate the number of moles of BaCl2 in 75 g using its molar mass.
The molar mass of BaCl2 can be calculated by summing the atomic masses of its elements:
Ba (barium) has an atomic mass of 137.33 g/mol and Cl (chlorine) has an atomic mass of 35.45 g/mol. Since there are two chlorine atoms in BaCl2, the molar mass of BaCl2 is:
Molar mass of BaCl2 = (1 * Ba) + (2 * Cl) = (1 * 137.33) + (2 * 35.45) = 208.23 g/mol
Now, we can calculate the number of moles:
Number of moles = mass / molar mass = 75 g / 208.23 g/mol ≈ 0.36 mol
Since the concentration of BaCl2 is 0.5 M (0.5 mol/L), we can use the formula for molarity to determine the volume of 0.5 M BaCl2 solution needed to get 0.36 mol:
Volume (L) = moles / concentration = 0.36 mol / 0.5 mol/L = 0.72 L
Finally, convert the volume from liters to milliliters:
Volume (mL) = 0.72 L * 1000 mL/L = 720 mL
Therefore, approximately 720 mL of 0.5 M BaCl2 are required to have 75 g of BaCl2.